# Homework Help: Complex Eigenvalues

1. Dec 13, 2004

### cepheid

Staff Emeritus
Hello:

-was solving for the eigenvalues of a matrix. Obtained:
$$\lambda = 1 \pm 2i$$

-substituted back into matrix to try and solve for the eigenvectors:

$$\left(\begin{array}{cc}2-2i & -2\\4 & -2-2i\end{array}\right) \left(\begin{array}{cc}x_1 \\ x_2 \end{array}\right) = \mathbf{0}$$

I'm having trouble solving this type of system. I'm just wondering what the general strategy is for tackling these types of matrices.

Thanks.

2. Dec 13, 2004

### shmoe

The same procedures as normal will work, don't let the complex numbers intimidate you! Divide the first row by 2-2i and go from there. To make (-2)/(2-2i) nicer to handle, you can multiply the numerator and denominator by the conjugate, 2+2i.

Once in reduced form, are you able to pick out the eigenvectors?

3. Dec 14, 2004

### cepheid

Staff Emeritus
I think I'm going nuts. The matrix I gave was for the eigenvalue 1+2i btw. Here's what I did:

divide row 1 by 2-2i. Then:

$$\frac{-2}{2-2i} \times \frac{2+2i}{2+2i} = \frac{-4-4i}{4-4i^2} = \frac{-4-4i}{4-4(-1)} = \frac{-2-2i}{4}$$

The matrix is now:

$$\left(\begin{array}{cc}1 & (-2-2i)/4 \\4 & -2-2i\end{array}\right) \left(\begin{array}{cc}\xi_1 \\ \xi_2 \end{array}\right) = \mathbf{0}$$

Both rows clearly represent the same equation, namely:

$$\xi_1 = \frac{1+i}{2}\xi_2$$

To me, that means:

$$\mathbf{\xi^{(1)}} = \left(\begin{array}{cc} \frac{1+i}{2} \\ 1 \end{array}\right)$$

But according to Boyce & diPrima:

$$\mathbf{\xi^{(1)}} = \left(\begin{array}{cc} 1 \\ 1-i \end{array}\right)$$

4. Dec 14, 2004

### cepheid

Staff Emeritus
Dammit! Nevermind. Those two answers are equivalent!!! I just had to go through a bunch more rigmarole to get it in their form! Express the ratio between xi1 and xi2 in the reverse way, and multiply by the complex conjugate yet again! (*wonders if there is some way to arrive at it in their form more directly*)

5. Dec 14, 2004

### Galileo

Instead of dividing row 1 by 2-2i, it may be easier to multiply by the right amount.

Muliply by 1+i first. This gives (2-2i)(1+i)=4 and -2(1+i)=-2-2i.
Dividing by 4, you can see immediately that the reduced form of the matrix is:

$$\left(\begin{array}{cc}1 & -\frac{1+i}{2}\\0 & 0\end{array}\right)$$

So $\frac{1+i}{2} \choose 1$ is solution. Rewrite like you did to get the other form.

If you want the 1 on top right away, you want a 1 in the upper right corner.
Start by subtracting the 1st row from the 2nd. You get 2+2i and -2i.
Multiply 2nd row by -i, you get: 2-2i and -2.
Dividing the first row by 2, you get:
$$\left(\begin{array}{cc}1-i & -1\\0 & 0\end{array}\right)$$

So a solution is: $1 \choose 1-i$.

Row reducing complex is usually tedious.

6. Dec 14, 2004

### dextercioby

Frien,there's no need to "go nuts"!! :tongue2: Try to put $x_{1} =1$ in the matrix equation and from the first equation u'll find $x_{2}=1-i$.If u put $x_{2} =1$ in the matrix equation and solving from the second equation u'll find your solution $x_{1}=\frac{1+i}{2}$.U have 2 vectors.See whether they are linear independent.They're not.So,it's irrelevant which u chose in the final solution;

Daniel.